# How do you factor completely x^3 – 2x^2 – 4x – 8?

Nov 4, 2016

${x}^{3} - 2 {x}^{2} - 4 x - 8 = \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) \left(x - {x}_{3}\right)$

where ${x}_{1} = \frac{2}{3} \left(1 + \sqrt[3]{19 + 3 \sqrt{33}} + \sqrt[3]{19 - 3 \sqrt{33}}\right)$

and ${x}_{2} , {x}_{3}$ are related Complex zeros...

#### Explanation:

$f \left(x\right) = {x}^{3} - 2 {x}^{2} - 4 x - 8$

$\textcolor{w h i t e}{}$
Discriminant

The discriminant $\Delta$ of a cubic polynomial in the form $a {x}^{3} + b {x}^{2} + c x + d$ is given by the formula:

$\Delta = {b}^{2} {c}^{2} - 4 a {c}^{3} - 4 {b}^{3} d - 27 {a}^{2} {d}^{2} + 18 a b c d$

In our example, $a = 1$, $b = - 2$, $c = - 4$ and $d = - 8$, so we find:

$\Delta = 64 + 256 - 256 - 1728 - 1152 = - 2816$

Since $\Delta < 0$ this cubic has $1$ Real zero and $2$ non-Real Complex zeros, which are Complex conjugates of one another.

$\textcolor{w h i t e}{}$
Tschirnhaus transformation

To make the task of solving the cubic simpler, we make the cubic simpler using a linear substitution known as a Tschirnhaus transformation.

$0 = 27 f \left(x\right) = 27 {x}^{3} - 54 {x}^{2} - 108 x - 216$

$= {\left(3 x - 2\right)}^{3} - 48 \left(3 x - 2\right) - 304$

$= {t}^{3} - 48 t - 304$

where $t = \left(3 x - 2\right)$

$\textcolor{w h i t e}{}$
Cardano's method

We want to solve:

${t}^{3} - 48 t - 304 = 0$

Let $t = u + v$.

Then:

${u}^{3} + {v}^{3} + 3 \left(u v - 16\right) \left(u + v\right) - 304 = 0$

Add the constraint $v = \frac{16}{u}$ to eliminate the $\left(u + v\right)$ term and get:

${u}^{3} + \frac{4096}{u} ^ 3 - 304 = 0$

Multiply through by ${u}^{3}$ and rearrange slightly to get:

${\left({u}^{3}\right)}^{2} - 304 \left({u}^{3}\right) + 4096 = 0$

Use the quadratic formula to find:

${u}^{3} = \frac{304 \pm \sqrt{{\left(- 304\right)}^{2} - 4 \left(1\right) \left(4096\right)}}{2 \cdot 1}$

$= \frac{304 \pm \sqrt{92416 - 16384}}{2}$

$= \frac{304 \pm \sqrt{76032}}{2}$

$= \frac{304 \pm 48 \sqrt{33}}{2}$

$= 152 \pm 24 \sqrt{33}$

$= {2}^{3} \cdot \left(19 \pm 3 \left(33\right)\right)$

Since this is Real and the derivation is symmetric in $u$ and $v$, we can use one of these roots for ${u}^{3}$ and the other for ${v}^{3}$ to find Real root:

${t}_{1} = 2 \sqrt[3]{19 + 3 \sqrt{33}} + 2 \sqrt[3]{19 - 3 \sqrt{33}}$

and related Complex roots:

${t}_{2} = 2 \omega \sqrt[3]{19 + 3 \sqrt{33}} + 2 {\omega}^{2} \sqrt[3]{19 - 3 \sqrt{33}}$

${t}_{3} = 2 {\omega}^{2} \sqrt[3]{19 + 3 \sqrt{33}} + 2 \omega \sqrt[3]{19 - 3 \sqrt{33}}$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive Complex cube root of $1$.

Now $x = \frac{1}{3} \left(2 + t\right) = \frac{2}{3} \left(1 + \frac{t}{2}\right)$. So the zeros of our original cubic are:

${x}_{1} = \frac{2}{3} \left(1 + \sqrt[3]{19 + 3 \sqrt{33}} + \sqrt[3]{19 - 3 \sqrt{33}}\right)$

${x}_{2} = \frac{2}{3} \left(1 + \omega \sqrt[3]{19 + 3 \sqrt{33}} + {\omega}^{2} \sqrt[3]{19 - 3 \sqrt{33}}\right)$

${x}_{3} = \frac{2}{3} \left(1 + {\omega}^{2} \sqrt[3]{19 + 3 \sqrt{33}} + \omega \sqrt[3]{19 - 3 \sqrt{33}}\right)$